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LINEAR EQUATIONS IN TWO VARIABLES
Here we have explained how to solve linear equations in two variables algebraically
and graphically. At the end of the explanation there are some practise questions.
HOW TO SOLVE LINEAR EQUATIONS IN TWO VARIABLES EXPLAINED
ELIMINATION METHOD TO SOLVE A SYSTEM OF EQUATIONS
We will explain this method using an example:
Example
( )
( )
3 2 18 1
4 3 2
x y
x y
-
+ -10
=
=
Multiply both sides of equation (1) by coefficient of x in second equation. Here multiply both
sides of eq. (1) by 4.
Multiply both sides of equation (2) by coefficient of x in first equation. Here multiply both
sides of eq. (2) by 3.
4(3x - 2y) = 4(18) (1)
3(4x + 3y) = 3(-10) (2)
12 8 72
12 9 30
- =
+ = -
x y
x y
Multiply the second equation by -1 and then add the two equations.
( )
( )
12 8 72 1
12 9 2
x y
x y
-
- -
=
= 30
__________________________________
-17 y =102
102
6
17
y = - = -
Now substitute -6 for y in either of the original equations.
3x - 2y =18
3x - 2y = 18
3x +12 = 18
3x = 6
x = 2
The solution of the set of equations is (2,-6).
You can check the solution by substituting into the second equation.
The point (2,-6) should make both the equations true then the solution is correct.
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SUBSTITUTION METHOD TO SOLVE LINEAR EQUATIONS
We will understand the method with an example.
Solve either of the equations for x or y.
7x -15y = 2 (1)
x + 2y = 3 (2)
Picking equation (2) to solve for x, x = 3 - 2y. Then substitute 3 - 2y in the first equation.
7x -15y = 2
7(3 - 2y) -15y = 2
-29y = -19
19
29
y =
Now substitute
19
29
for y in either of the original equations and solve for x.
x + 2y = 3
19
2 3
29
x
  +   =
 
38
3
29
x + =
38 87 38
3
29 29
x
= - = -
49
29
x =
The solution is
49 19
,
29 29
 
 
 
.
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SOLVING LINEAR EQUATIONS IN TWO VARIABLES GRAHICALLY
A system of equations can have a single or unique solution, no solution or infinitely many
solutions.
Let us look at each case one by one:
1. Look at the following two equations:
First step is to write the two equations in slope-intercept form by solving for y.
) 2 0
)3 4 20
i x y
ii x y
- =
+ =
2 0
2
1
2
x y
y x
y x
- =
=
=
Solve for y
Now the line is in the form y=mx+c
Slope of the line is +1/2 and its y intercept is zero
3 4 20
4 3 20
3 20
4 4
3
5
4
x y
y x
y x
y x
+ =
= - +
= - +
= - +
Solve for y
Now the line is in the form y=mx+c
Slope of the line is -3/4 and its y intercept is +5
As the two lines have different slopes , the graphs of the two equations are intersecting
lines.
Second step Graph each equation
Graph the equation y=1/2x
x-intercept y-intercept Another point
y=0
1/2x=0
x=0
x=0
y=1/2(0)
y=0
x=1
y=1/2(1)
y=1/2
( 0,0) (0,0) (1,1/2)
This gives us two distinct points (0,0) and (1,1/2) so that we can graph the equation.
Next we graph the second equation y = -3/4x+5
x-intercept y-intercept Another point
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y=0
0 = -3/4x+5
3/4x = 5
x=20/3
x=0
y = -3/4(0)+5
y = 5
x=4
y = -3/4(4)+5
y = -3+5
y = 2
(20/3,0)
(0,5)
(4,2)
Note: we wanted to avoid a fraction as they are difficult to plot, so we chose x as r. we could
have taken x as some other multiple of 4 also.
This gives us three points (20/3, 0), (0, 5) and (4,2) to graph the equation y = -3/4x+5.
Drawing both the graphs on the same paper shows
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The point (4,2) is a solution of the system of equations as it lies on both the graphs.
2. Next we have another example.
The given set of equations is
i) 2x + 3y = 9
ii) 4x + 6y = 18
Writing first equation in slope intercept form we get
2 3 9
3 2 9
2 9
3 3
2
3
3
x y
y x
y x
y x
+ =
= - +
= - +
= - +
Solve for y
Now the line is in the form y=mx+c
Slope of the line is -2/3 and its y intercept is 3
Similarly writing second equation in slope intercept form we get
4 6 18
6 4 18
4 18
6 6
2
3
3
x y
y x
y x
y x
+ =
= - +
= - +
= - +
Solve for y
Now the line is in the form y=mx+c
Slope of the line is -2/3 and its y intercept is 3
These shows both the equations are identical. Graphs for both equations will be the same
line. These equations have infinite number of solutions as each point on the graph satisfies
both equations.
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Graphing first and second equation we get
x-intercept y-intercept Another point
y=0
0 = -2/3x+3
2/3x = 3
x=9/2
x=4.5
x=0
y = -2/3 (0)+3
y = 3
x =3
y = -2/3(3)+3
y = -2+3
y = 1
(4.5,0)
(0,3)
(3,1)
Using these 3 points we can graph the equations
Each point on the graph is a solution of both equations. The given set of equations have
infinite number of solutions.
3. Let us look at yet another example
i) x + 2y - 4 = 0
ii) 2x + 4y -12 = 0
First the two equations can be written in slope intercept form as follows.
2 4 0
2 4
1 4
2 2
1
2
2
x y
y x
y x
y x
+ - =
= - +
= - +
= - +
Solve for y
Now the line is in the form y=mx+c
Slope of the line is -1/2 and its y intercept is 2
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2 4 12 0
4 2 12
2 12
4 4
1
3
2
x y
y x
y x
y x
+ - =
= - +
= - +
= - +
Solve for y
Now the line is in the form y=mx+c
Slope of the line is -1/2 and its y intercept is 3
Here both lines have the same slope, but their y intercepts are different. This means when
you graph the equations, the lines will be parallel to each other.
Graphing the two equations we get-
These lines will not intersect at any point. So the given set of equations has no solution, in
other words, there is no point that satisfies both the equations simultaneously.
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Questions for Practice
1. Solve the following system of simultaneous linear equations graphically.
(i) 3x + 2y = 8
2 46
5x y
3 3
- =
(ii) x - 2y = 5
2x – 4y = 10
2. Solve the following system of equations:
(i) 2x - 3y = 0
5x + 2y = 0
(ii)
x
y 8
2
+ =
7
10
y
x
2
=
+
3. Solve the following equations in x and y using cross-multiplication method
(i)
( )2
2 2
a + b
ax by 1, bx+ay = 1
a b
+ = -
+
(ii) ( - ) + ( + ) = 2 - - 2 a b x a b y a 2ab b
( ) ( ) 2 2 a + b x + y = a + b
4. In each of the following system of equations find if the system has a unique solution,
no solution or infinitely many solutions.
(i) 4x + 7y = 10
35
10x y 25
2
+ =
(ii) 4y - 3x = 5
9 15
x 6y
2 2
- =
5. Find the values of k for which the system has (i) a unique solution (ii) no solution.
kx 2y 5
3x y 1
+ =
+ =
6. Find the value of c for which the system has infinitely many solutions.
cx 3y c 3
12x cy c
+ = -
+ =
7. In a 0 0 0 ABC, ÐA = x ,ÐB = 3x and ÐC = y
If 3y - 5x = 30, prove that the triangle is right angled.
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8. A and B have certain number of toffees. A says to B, “if you give 30 of your toffees I will
have twice as many as left with you.” B replies “If you give me 10, I will have twice as
many as left with you”. How many toffees does each have?
9. Seven years ago the age of the father was 7 times the age of his son; three years hence,
the age of the father will be 3 times the age of the son. Find their present ages.
10. A farmer sold a calf and a cow for Rs. 760, thereby making a profit of 25% on the calf
and 10% on the cow. By selling them for Rs. 767.50, he would have realised a profit of
10% on the calf and 25% on the cow. Find cost of each.
11. For what value of k, the following system of equations:
3x 4y 6
6x 8y k
+ =
+ = represent coincident lines.
12. In the △ABC, one of the angles is 50% more than the sum of the other two angles. Find
the largest angle.
13. Solve the system of equations
( )
4 6
5
x 3 y 4
5 3
1 x 3, y 4
x 3 y 4
+ =
- -
- = ¹ ¹
- -
14. Students of a class are made to stand in rows. If one student is extra in a row, there
would be 2 rows less. If one student is less in a row there would be 3 rows more. Find
the number of students in the class.
15. Solve the system of equations by the method of cross-multiplication.
2 2 2
2 2 2
a x b y c
b x a y d
+ =
+ =
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ANSWERS
1. (i) x = 3, y = –½
(ii) Every solution of x – 2y = 5 is a solution.
Infinite Solutions.
2. (i) x = 0, y = 0
(ii) -22/5, 51/5
3. (i) =
2 + 2 2 + 2
a b
x , y =
a b a b
(ii) x = a + b ,
2ab
y
a b
= -
+
4. (i) infinitely many solutions
(ii) no solution
5. (i) k ¹ 6
(ii) k = 6
6. c = 6
7. 300, 600, 900
8. 34, 62
9. 42 yrs, 12 yrs
10. Cow = 519, calf = 151
11. k = 12
12. 1080
13. 5, 6
14. 60
15.
2 2 2 2
4 4
a c b d
x
a b
= -
-
,
2 2 2 2
4 4
a d b c
y
a b
= -
-